$\lim_{x\to\infty}\dfrac{x^4}{3x^2-7x}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{3}$ (Choice C) C $\dfrac{2}{3}$ (Choice D) D $\infty$
Explanation: $\lim_{x\to\infty} x^4=\infty$ and $\lim_{x\to\infty} 3x^2-7x=\infty$, so $\lim_{x\to\infty}\dfrac{x^4}{3x^2-7x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{x^4}{3x^2-7x} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[x^4\right]}{\dfrac{d}{dx}[3x^2-7x]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{4x^3}{6x-7} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[4x^3\right]}{\dfrac{d}{dx}[6x-7]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{12x^2}{6} \\\\ &=\infty \end{aligned}$ Note that we used l'Hôpital's rule twice, because the first time we used it, we ended with the indeterminate form $\dfrac{\infty}{\infty}$ too. Also note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[4x^3\right]}{\dfrac{d}{dx}[6x-7]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{x^4}{3x^2-7x}=\infty$.